R 用最新的非 NA 值替换 NA

在data.frame(或data.table)中，我想用最接近的前一个非NA值"向前填充"NAs.下面是一个使用向量(而不是data.frame)的简单示例:

> y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)

我想要一个函数fill.NAs()，它允许我构造yy，这样:

> yy
[1] NA NA NA  2  2  2  2  3  3  3  4  4

我需要对许多(总计~1 Tb)小型data.frames(~30-50MB)重复此操作，其中一行是NA，所有条目都是NA.解决这个问题的好方法是什么？

我设计的丑陋解决方案使用了以下功能:

last <- function (x){
    x[length(x)]
}    

fill.NAs <- function(isNA){
if (isNA[1] == 1) {
    isNA[1:max({which(isNA==0)[1]-1},1)] <- 0 # first is NAs 
                                              # can't be forward filled
}
isNA.neg <- isNA.pos <- isNA.diff <- diff(isNA)
isNA.pos[isNA.diff < 0] <- 0
isNA.neg[isNA.diff > 0] <- 0
which.isNA.neg <- which(as.logical(isNA.neg))
if (length(which.isNA.neg)==0) return(NULL) # generates warnings later, but works
which.isNA.pos <- which(as.logical(isNA.pos))
which.isNA <- which(as.logical(isNA))
if (length(which.isNA.neg)==length(which.isNA.pos)){
    replacement <- rep(which.isNA.pos[2:length(which.isNA.neg)], 
                                which.isNA.neg[2:max(length(which.isNA.neg)-1,2)] - 
                                which.isNA.pos[1:max(length(which.isNA.neg)-1,1)])      
    replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
} else {
    replacement <- rep(which.isNA.pos[1:length(which.isNA.neg)], which.isNA.neg - which.isNA.pos[1:length(which.isNA.neg)])     
    replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
}
replacement
}

功能fill.NAs的使用如下:

y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)
isNA <- as.numeric(is.na(y))
replacement <- fill.NAs(isNA)
if (length(replacement)){
which.isNA <- which(as.logical(isNA))
to.replace <- which.isNA[which(isNA==0)[1]:length(which.isNA)]
y[to.replace] <- y[replacement]
} 

100

> y
[1] NA  2  2  2  2  3  3  3  4  4  4

... 这似乎有效.但是，老兄，它丑吗！有什么建议吗？